# Partial Fraction Expansion with repeated & complex roots

1. Aug 23, 2009

### CremeBrulee

I am having trouble to find a straight forward method for finding coefficients using partial fraction expansion with repeated or complex roots. My study notes arent too clear so I am finding the differentiation method hard to follow for repeated roots. As for complex roots I can find the roots using quadratic formula but then not sure how to find the coefficients? Are there any simple methods out there that can used used for these types of equations?Any help would be appriciated?

Repeated roots
4/s(s+1)^2

4/s(s+1)^2=k1/s + k2/s+1 + k3/(s+1)^2

k1=4; k3=-4 (Using cover up method)

k2=??

Complex roots
2/s(s^2+s+1)

(s^2+s+1) = -1/2 +/- j sqrt(3/2)

2. Aug 23, 2009

### Feldoh

That "cover-up" method doesn't work with any polynomial with a degree higher than 2.

You'll have to solve k1/s + k2/s+1 + k3/(s+1)^2 the normal way.

Last edited: Aug 23, 2009
3. Aug 23, 2009

### coomast

CremeBrulee,

I do not know what exactly is meant by the cover-up method, but these two examples are very straightforward. You start very well, but than you seem to be lost. I will explain the first one, the second one you need to do yourself. (I wrote the first one down on a piece of paper in about 20 seconds.... the typing is a way bit longer....)

OK, you have:

$$\frac{4}{s(s+1)^2}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{(s+1)^2}$$

Or:

$$A(s+1)^2+Bs(s+1)+Cs=4$$
$$As^2+2As+A+Bs^2+Bs+Cs=4$$
$$(A+B)s^2+(2A+B+C)s+A=4$$

setting the LHS and the RHS equal to each other in the different s-terms we have already:

$$A=4$$

$$A+B=0$$

thus:

$$B=-4$$

$$2A+B+C=0$$

thus:

$$C=-4$$

the equation becomes now:

$$\frac{4}{s(s+1)^2}=\frac{4}{s}+\frac{-4}{s+1}+\frac{-4}{(s+1)^2}$$

Can you do the second one in the same manner? Do not try to find the roots of the quadratic that is not necessary. Write down the first step and in case you are not sure how to split it up, come back and ask.

coomast

4. Aug 23, 2009

### HallsofIvy

Staff Emeritus
As Coomast said, you can quickly arrive at
$$A(s+1)^2+Bs(s+1)+Cs=4$$

You can simplify a lot by taking s= -1:
$$A(0)^2+ B(-1)(0)+ C(-1)= 4$$
so that you get C= -4.

You can simplify a lot by taking s= 0:
$$A(0+1)^2+ B(0)(0+1)+ C(0)= 4$$
so that you get A= 4.

Finally, just choose some third number: s= 1 will do.
$$A(1+1)^2+ B(1)(1+1)+ C(1)= 4$$
$$4A+ 2B+ C= 4$$

And, since A= 4 and C= -4,
$$16+ 2B- 4= 4$$
so that
$$2B= -8$$
and B= -2.

5. Aug 24, 2009

### CremeBrulee

Thanks for the replies everyone. This process seems really straightforward but I'm still having trouble...grrr.

I do have a couple of questions:

How do I determine how to correctly split up the equation ie what s term goes next to each coefficient A, B , C, etc? I tried the second question but wasn't sure how to split it? What is the convention when splitting both repeated and complex roots?

The way I understood the explanation is that the s terms in brackets equal zero and the singular A equals 4.

$$(A+B)s^2+(2A+B+C)s+A=4$$

Then using the value found for A it can be plugged into the remaining coeffs in the s terms in brackets that equal zero to find remaining coefficients. Are the s terms in brackets always zero?

$$A+B=0$$ ; $$2A+B+C=0$$

But what happens if there isn't a singular coefficient or say if there were two repeated roots and so two singular values? Does this process still work?

I also tried the method using Hallsof Ivy suggested (s=0, s=1, s=-1) but I found B=-8? However, I know the answer is A=4, B=-4 & C=-4 from my text book, so we were both incorrect?